let f(u)=e^u
and u= x^3
f'(u)=e^u du/dx
but du/dx= 3x^2
f'(x)=e^(x^3) * 3x^2
let f be the function given by f(x)=-2e^x^3. F0r what value of x is the slope of the line tangent to the graph of f at (x,f(x)) equal to -6
2 answers
so, now you just want to find where
-2*3x^2 e^x^3 = -6
x^2 e^x^3 = 1
I think a graphical solution will be easiest.
-2*3x^2 e^x^3 = -6
x^2 e^x^3 = 1
I think a graphical solution will be easiest.